Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in PYTHON):

Solution:
n, m = map(int, raw_input().split())
a, b = [], []
for _ in range(n):
    a.append(map(int, raw_input().split()))
    b.append(a[_][:])
r, c = [], []
sumRC = 0
for i in range(n):
    tmp = min(a[i])
    for j in range(m): a[i][j] -= tmp
    for j in range(tmp): r.append(i + 1)
for i in range(m):
    tmp = min([a[j][i] for j in range(n)])
    for j in range(n): a[j][i] -= tmp
    for k in range(tmp): c.append(i + 1)
for i in range(n):
    for j in range(m):
        sumRC += a[i][j]

rC, cC = [], []
sumCR = 0
for i in range(m):
    tmp = min([b[j][i] for j in range(n)])
    for j in range(n): b[j][i] -= tmp
    for k in range(tmp): cC.append(i + 1)
for i in range(n):
    tmp = min(b[i])
    for j in range(m): b[i][j] -= tmp
    for j in range(tmp): rC.append(i + 1)
for i in range(n):
    for j in range(m):
        sumCR += b[i][j]
if sumRC == 0 and sumCR == 0:
    if len(r) + len(c) <= len(rC) + len(cC): R, C = r, c
    else: R, C = rC, cC
    print len(R) + len(C)
    for i in R:
        print "row " + str(i)
    for i in C:
        print "col " + str(i)
elif sumRC == 0:
    R, C = r, c
    print len(R) + len(C)
    for i in R:
        print "row " + str(i)
    for i in C:
        print "col " + str(i)
elif sumCR == 0:
    R, C = rC, cC
    print len(R) + len(C)
    for i in R:
        print "row " + str(i)
    for i in C:
        print "col " + str(i)
else:
    print -1